Trigonometry

Trigonometry

Easy Questions
Moderate Questions
Difficult Questions
Trigonometry Tutorial

Aptitude

Simplification
Number System
Fractions
Elementary Algebra
LCM and HCF
Approximation
Unitary Method
Linear Equation
Quadratic Equation
Discount
Average
Surds and Indices
Percentage
Square root and cube root
Order of Magnitude
Work and Wages
Odd Man Out and Series
Algebra
Profit and Loss
Stocks and Shares
True Discount
Ratio, Proportion
Partnership
Alligation or Mixture
Time and Work
Pipes and Cistern
Speed, Time and Distance
Problem on Trains
Height and Distance
Banker's Discount
Boats and Streams
Races and games
Problems on Ages
Clocks and Calendars
Simple interest
Compound Interest
Sets and Functions
Area and Perimeter
Volume and Surface Area of Solid Figures
Sequences and Series
Plane Geometry
Logarithm
Probability
Permutation and Combination
Statistics
Mensuration
Trigonometry
Aptitude miscellaneous
  1. If tan α = 2, then the value of
    sinα
    		is
    sin3α + cos3α








  1. View Hint View Answer Discuss in Forum

    tanα = 2
    ∴  sec2α = 1 + tan2α = 1 + 22
    = 1 + 4 = 5

    Expression =
    sinα
    sin3α + cos3α

    =
    sinα
    cos3α
    sin3α
    		 +
    cos3α
    cos3αcos3α

    =
    sinα
    		 .
    1
    		 .
    1
    cosα cos2α (tan3α + 1)

    = tanα.sec2α.
    1
    (tan3α + 1)

    = 2 × 5 ×
    1
    (23 + 1)

    =
    10
    9

    Correct Option: C

    tanα = 2
    ∴  sec2α = 1 + tan2α = 1 + 22
    = 1 + 4 = 5

    Expression =
    sinα
    sin3α + cos3α

    =
    sinα
    cos3α
    sin3α
    		 +
    cos3α
    cos3αcos3α

    =
    sinα
    		 .
    1
    		 .
    1
    cosα cos2α (tan3α + 1)

    = tanα.sec2α.
    1
    (tan3α + 1)

    = 2 × 5 ×
    1
    (23 + 1)

    =
    10
    9

  1. If √2 tan2θ = √6 and 0° < θ < 45°, then the value of sinθ + √3cosθ – 2tan2θ is








  1. View Hint View Answer Discuss in Forum

    √2 tan2θ = √6

    ⇒  tan2θ =
    √6
    		 = √3
    √2

    ⇒  tan2θ = tan 60°
    ⇒  2θ = 60° ⇒ θ = 30°
    ∴  sinθ + √3 cosθ – 2 tan2θ
    = sin 30° + √3 cos 30° – 2 tan230°
    =
    1
    		 + √3 ×
    √3
    		 – 2
    1
    		2
    22√3

    =
    1
    		 +
    3
    		 −
    2
    223

    =
    3 + 9 − 4
    		 =
    8
    		 =
    4
    663

    Correct Option: B

    √2 tan2θ = √6

    ⇒  tan2θ =
    √6
    		 = √3
    √2

    ⇒  tan2θ = tan 60°
    ⇒  2θ = 60° ⇒ θ = 30°
    ∴  sinθ + √3 cosθ – 2 tan2θ
    = sin 30° + √3 cos 30° – 2 tan230°
    =
    1
    		 + √3 ×
    √3
    		 – 2
    1
    		2
    22√3

    =
    1
    		 +
    3
    		 −
    2
    223

    =
    3 + 9 − 4
    		 =
    8
    		 =
    4
    663

  1. Find the value of
    sin 27°
    		2 +
    cos 63°
    		2
    cos 63°sin 27°








  1. View Hint View Answer Discuss in Forum

    sin 27° = sin (90° – 63°) = cos 63°
    [∵  sin (90° – θ) = cosθ]

    ∴ 
    sin 27°
    		2 +
    cos 63°
    		2
    cos 63°sin 27°

    =
    sin 27°
    		2 +
    sin 27°
    		2
    sin 27°sin 27°

    = 1 + 1 = 2

    Correct Option: B

    sin 27° = sin (90° – 63°) = cos 63°
    [∵  sin (90° – θ) = cosθ]

    ∴ 
    sin 27°
    		2 +
    cos 63°
    		2
    cos 63°sin 27°

    =
    sin 27°
    		2 +
    sin 27°
    		2
    sin 27°sin 27°

    = 1 + 1 = 2

  1. The value of tan80° tan10° + sin2 70° + sin270° + sin220° is








  1. View Hint View Answer Discuss in Forum

    tan 80° . tan 10° + sin270° + sin220°
    = tan (90° – 10°) . tan 10° + sin2(90° – 20°) + sin220°
    = cot 10° . tan 10° + cos220° + sin220°
    = 1 + 1 = 2
    [∵  sin (90° – θ) = cosθ; tan (90° – θ) = cotθ; tanθ . cotθ = 1]

    Correct Option: C

    tan 80° . tan 10° + sin270° + sin220°
    = tan (90° – 10°) . tan 10° + sin2(90° – 20°) + sin220°
    = cot 10° . tan 10° + cos220° + sin220°
    = 1 + 1 = 2
    [∵  sin (90° – θ) = cosθ; tan (90° – θ) = cotθ; tanθ . cotθ = 1]